Implications of the two Functor axioms As I am very new to category theory, please let me know if I misspeak. 
Let C and D be categories, F a functor from C to D, and f a morphism in Hom$_C$(X, Y). From what I can understand, the functor axioms (pasted below) imply that F(f) lives in Hom$_D$(F(X), F(Y)).
i) F(g$\circ$f) = F(g)$\circ$F(f) for all morphisms f : X $\rightarrow$ Y and g : Y $\rightarrow$ Z in Map(C).
ii) F(id$_X$) = id$_{F(X)}$ for all objects, X, in Ob(C).
So far, where F(f) must live, is the only fact I know about functors. I was wondering whether or not there are examples of two distinct objects being mapped (by a functor) to a single object. I tried to use the axioms to get some sort of resolution, but made no progress. 
If anyone has any interesting facts about functors, or knows of an example to what was outlined above, I would greatly appreciate it!
 A: I would comment but I do not yet have enough reputation, so I decided to develop my point a bit more:
Apart from what has been said in the comments about functors that map different objects to the same object, you asked about implications of the functor axioms. Now one certainly can derive that $F(f)$ must be an arrow $F(a) \rightarrow F(b)$ for $f:a \rightarrow b$, however, this is not the whole point of them. Instead, they serve to ensure that functors preserve structure in a certain sense.
Indeed, $F(g \circ f) = F(g) \circ F(f)$ ensures that $F$ preserves commuting triangles:
\begin{array}{ccc}
a & \xrightarrow{f} & b  \\
 & \searrow & \downarrow{\scriptstyle g} \\
  & & c
\end{array}
is mapped to
\begin{array}{ccc}
F(a) & \xrightarrow{F(f)} & F(b)  \\
 & \searrow & \downarrow{\scriptstyle F(g)} \\
  & & F(c)
\end{array}
so if the first diagram commutes, it would be necessary for $g \circ f$ to be mapped to $F(g) \circ F(f)$ to ensure that the second diagram also commutes.
Now $\newcommand{\id}{\text{id}} F(\id_a) = \id_{F(a)}$ ensures that $F$ preserves isomorphisms: If $f: a \rightarrow b$ and $g: b \rightarrow a$ are inverses, i.e. $f \circ g = \text{id}_b$ and $g \circ f = \text{id}_a$, then we also want that $F(f)$ be inverse to $F(g)$, i.e. $F(g) \circ F(f) = F(g \circ f) = F(\text{id}_a)$ should be the identity on $F(a)$.
